Lemma 53.24.1. Let $k$ be a field. Let $c : X \to Y$ be a morphism of proper schemes over $k$ Assume
$\mathcal{O}_ Y = c_*\mathcal{O}_ X$ and $R^1c_*\mathcal{O}_ X = 0$,
$X$ and $Y$ are reduced, Gorenstein, and have dimension $1$,
$\exists \ m \in \mathbf{Z}$ with $H^1(X, \omega _ X^{\otimes m}) = 0$ and $\omega _ X^{\otimes m}$ generated by global sections.
Then $c^*\omega _ Y \cong \omega _ X$.
Proof. The fibres of $c$ are geometrically connected by More on Morphisms, Theorem 37.53.4. In particular $c$ is surjective. There are finitely many closed points $y = y_1, \ldots , y_ r$ of $Y$ where $X_ y$ has dimension $1$ and over $Y \setminus \{ y_1, \ldots , y_ r\} $ the morphism $c$ is an isomorphism. Some details omitted; hint: outside of $\{ y_1, \ldots , y_ r\} $ the morphism $c$ is finite, see Cohomology of Schemes, Lemma 30.21.1.
Let us carefully construct a map $b : c^*\omega _ Y \to \omega _ X$. Denote $f : X \to \mathop{\mathrm{Spec}}(k)$ and $g : Y \to \mathop{\mathrm{Spec}}(k)$ the structure morphisms. We have $f^!k = \omega _ X[1]$ and $g^!k = \omega _ Y[1]$, see Lemma 53.4.1 and its proof. Then $f^! = c^! \circ g^!$ and hence $c^!\omega _ Y = \omega _ X$. Thus there is a functorial isomorphism
for coherent $\mathcal{O}_ X$-modules $\mathcal{F}$ by definition of $c^!$1. This isomorphism is induced by a trace map $t : Rc_*\omega _ X \to \omega _ Y$ (the counit of the adjunction). By the projection formula (Cohomology, Lemma 20.54.2) the canonical map $a : \omega _ Y \to Rc_*c^*\omega _ Y$ is an isomorphism. Combining the above we see there is a canonical map $b : c^*\omega _ Y \to \omega _ X$ such that
In particular, if we restrict $b$ to $c^{-1}(Y \setminus \{ y_1, \ldots , y_ r\} )$ then it is an isomorphism (because it is a map between invertible modules whose composition with another gives the isomorphism $a^{-1}$).
Choose $m \in \mathbf{Z}$ as in (3) consider the map
This map is injective because $Y$ is reduced and by the last property of $b$ mentioned in its construction. By Riemann-Roch (Lemma 53.5.2) we have $\chi (X, \omega _ X^{\otimes m}) =\chi (Y, \omega _ Y^{\otimes m})$. Thus
and we conclude $b^{\otimes m}$ induces an isomorphism on global sections. So $b^{\otimes m} : c^*\omega _ Y^{\otimes m} \to \omega _ X^{\otimes m}$ is surjective as generators of $\omega _ X^{\otimes m}$ are in the image. Hence $b^{\otimes m}$ is an isomorphism. Thus $b$ is an isomorphism. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)